Question

A 0.40 kg block can slide up and down a rough a 10-m-high, 30-m-long slope. At the bottom, a stiff spring with spring constant 800 N/m is compressed 0.50 m and used to launch the block up the slope. The friction force on the block from the slope is 1.2 N. What is the speed of the block when it reaches the top of the slope, in m/s

Answer 1

Answer:

$v=11.135m/s$

Explanation:

From the question we are told that

Mass $m=0.40kg$

Frictional friction $f=1.2$

Spring constant $k=800N/m$

Compressed $x=0.50$

Generally the energy balance equation is mathematically given as

$1/2kx^2=f*30+m*g*sin \theta*30+1/2m*v^2$

if $(sin=10/30-1/3)$

$1/2(800)(0.5)^2=(1.2)*30+(0.4)*9.8*1/3*30+1/2(0.4)*v^2$

$100=36+39.2+0.20v^2$

$v^2=\frac{24.8}{0.20}$

$v=\sqrt{124}$

$v=11.135m/s$